What is the derivative of 3/(sqrt(16+x^4)) ?

The derivative of 3/(sqrt(16+x^4)) is -(6x^3)/((16+x^4)^{3/2)}

What is the first derivative of 3/(sqrt(16+x^4)) ?

The first derivative of 3/(sqrt(16+x^4)) is -(6x^3)/((16+x^4)^{3/2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (\frac{3}{16 + x ^{4}})

- \frac{6 x ^{3}}{( 16 + x ^{4} ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d x} (\frac{3}{16 + x ^{4}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d x} (\frac{1}{16 + x ^{4}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 3 \frac{d}{d x} ((16 + x^{4})^{- \frac{1}{2}})

Apply the chain rule:

- \frac{1}{2 ( 16 + x ^{4} ) ^{\frac{3}{2}}} \frac{d}{d x} (16 + x^{4})

= - \frac{1}{2 ( 16 + x ^{4} ) ^{\frac{3}{2}}} \frac{d}{d x} (16 + x^{4})

\frac{d}{d x} (16 + x^{4}) = 4 x^{3}

= 3 (- \frac{1}{2 ( 16 + x ^{4} ) ^{\frac{3}{2}}} \cdot 4 x^{3})

Simplify

3 (- \frac{1}{2 ( 16 + x ^{4} ) ^{\frac{3}{2}}} \cdot 4 x^{3}) : - \frac{6 x ^{3}}{( 16 + x ^{4} ) ^{\frac{3}{2}}}

= - \frac{6 x ^{3}}{( 16 + x ^{4} ) ^{\frac{3}{2}}}

\frac{\partial}{\partial θ} (2 cos (θ))

\int \frac{- 4}{x} d x

\int_{0}^{2} (x - 1) e^{(x - 1)^{2}} d x

f (x) = (2 x - 5)^{3}, at x = 1

\int e^{8 x} cos (3 x) d x

What is the derivative of 3/(sqrt(16+x^4)) ?
The derivative of 3/(sqrt(16+x^4)) is -(6x^3)/((16+x^4)^{3/2)}
What is the first derivative of 3/(sqrt(16+x^4)) ?
The first derivative of 3/(sqrt(16+x^4)) is -(6x^3)/((16+x^4)^{3/2)}